【POJ 2387】 Til the Cows Come Home 最短路 Dijkstra

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在一片广阔的牧场中，Bessie需要找到回谷仓的最短路径，以避免被农夫John发现并提前开始早晨的挤奶工作。面对复杂的地形和多个地标，Bessie必须利用现有的双向牛径来规划她的路线。本篇将介绍如何使用Dijkstra算法解决这个问题，通过遍历所有可能的路径，确定从苹果树林到谷仓的最短距离。

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Til the Cows Come Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 80865		Accepted: 26668

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N

* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

Sample Output

题意：给你两个数字T,N,有N个点，T条边，接下来T行，每行三个数字，u，v, w,代表从u点到v点的距离为w，问你从1点走到N点的最短距离为多少。

题解：一道最短路的水题。


#include<cstdio>
using namespace std;
const int maxn=1000+7;
const int INF=1<<29;
int cost[maxn][maxn];
int low[maxn];
int vis[maxn];
int t,n,u,v,w;
int Dijkstra()
{
    int min,pos;
    for(int i=0;i<n;i++)  vis[i]=0;
    for(int i=0;i<n;i++){
        if(i==0) low[i]=0;  //源点距离初始化为0
        else low[i]=INF;    //其余点距离初始化为无穷大
    }
    for(int i=0;i<n;i++){
        min=INF,pos=-1;
        for(int j=0;j<n;j++){
            if(vis[j]==0&&low[j]<min){
                min=low[j];
                pos=j;
            }
        }
        vis[pos]=1;
        for(int j=0;j<n;j++){      //松弛
            if(!vis[j]&&low[pos]+cost[pos][j]<low[j]){
                low[j]=low[pos]+cost[pos][j];
            }
        }
    }
    return low[n-1];
}
int main()
{
    while(scanf("%d %d",&t,&n)==2){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                cost[i][j]=INF;
            }
        }
        for(int i=0;i<t;i++){
            scanf("%d %d %d",&u,&v,&w);
            if(w<cost[u-1][v-1]){        //重边判断
                cost[u-1][v-1]=cost[v-1][u-1]=w;
            }
        }
        int res=Dijkstra();
        printf("%d\n",res);
    }
    return 0;
}